RS Agarwal Solution | Class 9th | Chapter-5 |  Coordinate Geometry | Edugrown

Exercise MCQ

Question 1

The equation of the x-axis is

(a) x = 0

(b) y = 0

(c) x = y

(d) x + y = 0Solution 1

Correct option: (b)

The equation of the x-axis is y = 0.Question 2

The equation of the y-axis is

(a) x = 0

(b) y = 0

(c) x = y

(d) x + y = 0Solution 2

Correct option: (a)

The equation of the y-axis is x = 0. Question 3

The point of the form (a,a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y = x

(d) the line x + y = 0Solution 3

Question 4

The point of the form (a,-a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y-x=0

(d) the line x + y = 0Solution 4

Question 5

The linear equation 3x – 5y = 15 has

(a) a unique solution

(b) two solutions

(c) infinitely many solutions

(d) no solutionSolution 5

Question 6

The equation 2x + 5y = 7 has a unique solution, if x and y are

(a) natural numbers

(b) rational numbers

(c) positive real numbers

(d) real numbersSolution 6

Correct option: (a)

The equation 2x + 5y = 7 has a unique solution, if x and y are natural numbers.

If we take x = 1 and y = 1, the given equation is satisfied. Question 7

The graph of y = 5 is a line

(a) making an intercept 5 on the x-axis

(b) making an intercept 5 on the y-axis

(c) parallel to the x-axis at a distance of 5 units from the origin

(d) parallel to the y-axis at a distance of 5 units from the originSolution 7

Correct option: (c)

The graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin. Question 8

The graph of x = 4 is a line

(a) making an intercept 4 on the x-axis

(b) making an intercept 4 on the y-axis

(c) parallel to the x-axis at a distance of 4 units from the origin

(d) parallel to the y-axis at a distance of 4 units from the originSolution 8

Correct option: (d)

The graph of x = 4 is a line parallel to the y-axis at a distance of 4 units from the origin. Question 9

The graph of x + 3 = 0 is a line

(a) making an intercept -3 on the x-axis

(b) making an intercept -3 on the y-axis

(c) parallel to the y-axis at a distance of 3 units to the left of y-axis

(d) parallel to the x-axis at a distance of 3 units below the x-axisSolution 9

Correct option: (c)

The graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis. Question 10

The graph of y + 2 = 0 is a line

(a) making an intercept -2 on the x-axis

(b) making an intercept -2 on the y-axis

(c) parallel to the x-axis at a distance of 2 units below the x-axis

(d) parallel to the y-axis at a distance of 2 units to the left of y-axisSolution 10

Correct option: (c)

The graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis. Question 11

The graph of the linear equation 2x + 3y = 6 meets the y-axis at the point

(a) (2, 0)

(b) (3, 0)

(c) (0, 2)

(d) (0, 3)Solution 11

Correct option: (c)

When a graph meets the y-axis, the x coordinate is zero.

Thus, substituting x = 0 in the given equation, we get

2(0) + 3y = 6

⇒ 3y = 6

⇒ y = 2

Hence, the required point is (0, 2).Question 12

The graph of the linear equation 2x + 5y = 10 meets the x-axis at the point

(a) (0, 2)

(b) (2, 0)

(c) (5, 0)

(d) (0, 5)Solution 12

Correct option: (c)

When a graph meets the x-axis, the y coordinate is zero.

Thus, substituting y = 0 in the given equation, we get

2x + 5(0) = 10

⇒ 2x = 10

⇒ x = 5

Hence, the required point is (5, 0). Question 13

The graph of the line x = 3 passes through the point

(a) (0,3)

(b) (2,3)

(c) (3,2)

(d) None of theseSolution 13

Question 14

The graph of the line y = 3 passes though the point

(a) (3, 0)

(b) (3, 2)

(c) (2, 3)

(d) none of theseSolution 14

Correct option: (c)

Since, the y coordinate is 3, the graph of the line y = 3 passes through the point (2, 3).Question 15

The graph of the line y = -3 does not pass through the point

(a) (2,-3)

(b) (3,-3)

(c) (0,-3)

(d) (-3,2)Solution 15

Question 16

The graph of the linear equation x-y=0 passes through the point

Solution 16

Question 17

If each of (-2,2), (0,0) and (2,-2) is a solution of a linear equation in x and y, then the equation is

(a) x-y=0

(b) x+y=0

(c) -x+2y=0

(d) x – 2y=0Solution 17

Question 18

How many linear equations can be satisfied by x = 2 and y = 3?

(a) only one

(b) only two

(c) only three

(d) Infinitely manySolution 18

Correct option: (d)

Infinitely many linear equations can be satisfied by x = 2 and y = 3. Question 19

A linear equation in two variable x and y is of the form ax+by+c=0, where

(a) a≠0, b≠0

(b) a≠0, b=0

(c) a=0, b≠0

(d) a= 0, c=0Solution 19

Question 20

If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is

(a) 6

(b) 5

(c) 2

(d) 4Solution 20

Correct option: (d)

Since, (2, 0) is a solution of the linear equation 2x + 3y = k, substituting x = 2 and y = 0 in the given equation, we have

2(2) + 3(0) = k

⇒ 4 + 0 = k

⇒ k = 4 Question 21

Any point on x-axis is of the form:

(a) (x,y), where x ≠0 and y ≠0

(b) (0,y), where y ≠0

(c) (x,0), where x ≠0

(d) (y,y), where y ≠0Solution 21

Question 22

Any point on y-axis is of the form

(a) (x,0), where x ≠ 0

(b) (0,y), where y ≠ 0

(c) (x,x), where x ≠ 0

(d) None of theseSolution 22

Question 23

x = 5, y = 2 is a solution of the linear equation

(a) x + 2y = 7

(b) 5x + 2y = 7

(c) x + y = 7

(d) 5x + y = 7Solution 23

Correct option: (c)

Substituting x = 5 and y = 2 in L.H.S. of equation x + y = 7, we get

L.H.S. = 5 + 2 = 7 = R.H.S.

Hence, x = 5 and y = 2 is a solution of the linear equation x + y = 7. Question 24

If the point (3, 4) lies on the graph of 3y = ax + 7 then the value of a is

(a) 

(b) 

(c) 

(d) Solution 24

Correct option: (b)

Since the point (3, 4) lies on the graph of 3y = ax + 7, substituting x = 3 and y = 4 in the given equation, we get

3(4) = a(3) + 7

⇒ 12 = 3a + 7

⇒ 3a = 5

Exercise Ex. 4B

Question 1(vii)

Draw the graph of each of the following equation.

y + 5 = 0 Solution 1(vii)

y + 5 = 0

⇒ y = -5, which is a line parallel to the X-axis, at a distance of 5 units from it, below the X-axis.

Question 1(viii)

Draw the graph of each of the following equation.

y = 4Solution 1(viii)

y = 4 is a line parallel to the X-axis, at a distance of 4 units from it, above the X-axis.

Question 1(i)

Draw the graph of each of the following equation.

x = 4Solution 1(i)

x = 4 is a line parallel to the Y-axis, at a distance of 4 units from it, to its right.

Question 1(ii)

Draw the graph of each of the following equation.

x + 4 = 0Solution 1(ii)

x + 4 = 0

⇒ x = -4, which is a line parallel to the Y-axis, at a distance of 4 units from it, to its left.

Question 1(iii)

Draw the graph of each of the following equation.

y = 3Solution 1(iii)

y = 3 is a line parallel to the X-axis, at a distance of 3 units from it, above the X-axis.

Question 1(iv)

Draw the graph of each of the following equation.

y = -3Solution 1(iv)

y = -3 is a line parallel to the X-axis, at a distance of 3 units from it, below the X-axis.

Question 1(v)

Draw the graph of each of the following equation.

x = -2Solution 1(v)

x = -2 is a line parallel to the Y-axis, at a distance of 2 units from it, to its left.

Question 1(vi)

Draw the graph of each of the following equation.

x = 5Solution 1(vi)

x = 5 is a line parallel to the Y-axis, at a distance of 5 units from it, to its right.

Question 2(i)

Draw the graph of the equation y = 3x.

From your graph, find the value of y when x = 2.Solution 2(i)

y = 3x

When x = 1, then y = 3(1) = 3

When x = -1, then y = 3(-1) = -3

Thus, we have the following table:

x	1	-1
y	3	-3

Now, plot the points A(1, 3) and B(-1, -3) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of y = 3x.

Reading the graph

Given: x = 2. Take a point M on the X-axis such that OM = 2.

Draw MP parallel to the Y-axis, cutting the line AB at P.

Clearly, PM = 6

Thus, when x = 2, then y = 6.Question 2(ii)

Draw the graph of the equation y = 3x. From your graph, find the value of y when x = -2.Solution 2(ii)

The given equation is y = 3x.

Putting x = 1, y = 3  1 = 3

Putting x = 2, y = 3  2 = 6

Thus, we have the following table:

x	1	2
y	3	6

Plot points (1,3) and (2,6) on a graph paper and join them to get the required graph.

Take a point P on the left of y-axis such that the distance of point P from the y-axis is 2 units.

Draw PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to x-axis meeting y-axis at N.

So, y = ON = -6.Question 3(ii)

Draw the graph of the equation x + 2y – 3 = 0.

From your graph, find the value of y when x = -5Solution 3(ii)

x + 2y – 3 = 0

⇒ 2y = 3 – x

When x = -1, then  

When x = 1, then 

Thus, we have the following table:

x	-1	1
y	2	1

Now, plot the points A(-1, 2) and B(1, 1) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + 2y – 3 = 0.

Reading the graph

Given: x = -5. Take a point M on the X-axis such that OM = -5.

Draw MP parallel to the Y-axis, cutting the line AB at P.

Clearly, PM = 4

Thus, when x = -5, then y = 4. Question 3(i)

Draw the graph of the equation x + 2y – 3 = 0. From your graph, find the value of y when x = 5.Solution 3(i)

The given equation is,

x + 2y – 3 = 0

x = 3 – 2y

Putting y = 1,x = 3 – (2 1) = 1

Putting y = 0,x = 3 – (2 0) = 3

Thus, we have the following table:

x	1	3
y	1	0

Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.

Take a point Q on x-axis such that OQ = 5.

Draw QP parallel to y-axis meeting the line (x = 3 – 2y) at P.

Through P, draw PM parallel to x-axis cutting y-axis at M.

So, y = OM = -1.Question 4

Draw the graph of the equation 2x – 3y = 5. From the graph, find (i) the value of y when x = 4, and (ii) the value of x when y = 3.Solution 4

The given equation is, 2x – 3y = 5

Now, if x = 4, then

And, if x = -2, then

Thus, we have the following table:

x	4	-2
y	1	-3

Plot points (4,1) and (-2,-3) on a graph paper and join them to get the required graph.

(i) When x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to its right cutting the line at Q and through Q draw a line parallel to x-axis cutting y-axis which is found to be at a distance of 1 units above x-axis.

Thus, y = 1 when x = 4.

(ii) When y = 3, draw a line parallel to x-axis at a distance of 3 units from x-axis and above it, cutting the line at point P. Through P, draw a line parallel to y-axis meeting x-axis at a point which is found be 7 units to the right of y axis.

Thus, when y = 3, x = 7.Question 5

Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.Solution 5

The given equation is 2x + y = 6

 y = 6 – 2x

Now, if x = 1, then y = 6 – 2  1 = 4

And, if x = 2, then y = 6 – 2  2 = 2

Thus, we have the following table:

x	1	2
y	4	2

Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.

We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.

So, the co-ordinates of P are (3,0).Question 6

Draw the graph of the equation 3x + 2y = 6. Find the coordinates of the point, where the graph cuts the y-axis.Solution 6

The given equation is 3x + 2y = 6

 2y = 6 – 3x

Now, if x = 2, then

And, if x = 4, then

Thus, we have the following table:

x	2	4
y	0	-3

Plot points (2, 0) and (4,-3) on a graph paper and join them to get the required graph.

We find that the line 3x + 2y = 6 cuts the y-axis at a point P which is 3 units above the x-axis.

So, co-ordinates of P are (0,3).Question 7

Draw the graphs of the equations 3x – 2y = 4 and x + y – 3 = 0. On the same graph paper, find the coordinates of the point where the two graph lines intersect.Solution 7

Graph of the equation 3x – 2y = 4

⇒ 2y = 3x – 4

When x = 2, then  

When x = -2, then 

Thus, we have the following table:

x	2	-2
y	1	-5

Now, plot the points A(2, 1) and B(-2, -5) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 3x – 2y = 4.

Graph of the equation x + y – 3 = 0

⇒ y = 3 – x

When x = 1, then y = 3 – 1 = 2 

When x = -1, then y = 3 – (-1) = 4

Thus, we have the following table:

x	1	-1
y	2	4

Now, plot the points C(1, 2) and D(-1, 4) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of x + y – 3 = 0.

The two graph lines intersect at point A(2, 1). Question 8(i)

Draw the graph of the line 4x + 3y = 24.

Write the coordinates of the points where this line intersects the x-axis and the y-axis.Solution 8(i)

4x + 3y = 24

⇒ 3y = 24 – 4x

When x = 0, then  

When x = 3, then 

Thus, we have the following table:

x	0	3
y	8	4

Now, plot the points A(0, 8) and B(3, 4) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 4x + 3y = 24.

Reading the graph

The graph of line 4x + 3y = 24 intersects the X-axis at point C(6, 0) and the Y-axis at point A(0, 8). Question 8(ii)

Draw the graph of the line 4x + 3y = 24.

Use this graph to find the area of the triangle formed by the graph line and the coordinate axes.Solution 8(ii)

4x + 3y = 24

⇒ 3y = 24 – 4x

When x = 0, then  

When x = 3, then 

Thus, we have the following table:

x	0	3
y	8	4

Now, plot the points A(0, 8) and B(3, 4) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 4x + 3y = 24.

Reading the graph

Required area = Area of ΔAOC

Question 9

Draw the graphs of the lines 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these two lines and the x-axis. Find the area of the shaded region.Solution 9

Graph of the equation 2x + y = 6

⇒ y = 6 – 2x

When x = 1, then y = 6 – 2(1) = 6 – 2 = 4 

When x = 2, then y = 6 – 2(2) = 6 – 4 = 2

Thus, we have the following table:

x	1	2
y	4	2

Now, plot the points A(1, 4) and B(2, 2) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 2x + y = 6.

Graph of the equation 2x – y + 2 = 0

⇒ y = 2x + 2

When x = -1, then y = 2(-1) + 2 = -2 + 2 = 0 

When x = 2, then y = 2(2) + 2 = 4 + 2 = 6

Thus, we have the following table:

x	-1	2
y	0	6

Now, plot the points C(-1, 0) and D(2, 6) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of 2x – y + 2 = 0.

The two graph lines intersect at point A(1, 4).

The area enclosed by the lines and X-axis is shown in the graph.

Draw AM perpendicular from A on X-axis.

PM = y-coordinate of point A(1, 4) = 4

And, CP = 4

Area of shaded region = Area of ΔACP

Question 10

Draw the graphs of the lines x – y = 1 and 2x + y = 8. Shade the area formed by these two lines and the y-axis. Also, find this area.Solution 10

Graph of the equation x – y = 1

⇒ y = x – 1

When x = 1, then y = 1 – 1 = 0 

When x = 2, then y = 2 – 1 = 1

Thus, we have the following table:

x	1	2
y	0	1

Now, plot the points A(1, 0) and B(2, 1) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x – y = 1.

Graph of the equation 2x + y = 8

⇒ y = 8 – 2x

When x = 2, then y = 8 – 2(2) = 8 – 4 = 4 

When x = 3, then y = 8 – 2(3) = 8 – 6 = 2 

Thus, we have the following table:

x	2	3
y	4	2

Now, plot the points C(2, 4) and D(3, 2) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of 2x + y = 8.

The two graph lines intersect at point D(3, 2).

The area enclosed by the lines and Y-axis is shown in the graph.

Draw DM perpendicular from D on Y-axis.

DM = x-coordinate of point D(3, 2) = 3

And, EF = 9

Area of shaded region = Area of ΔDEF

Question 11

Draw the graph for each of the equations x + y = 6 and x – y = 2 on the same graph paper and find the coordinates of the point where the two straight lines intersect.

*Back answer incorrect.Solution 11

Graph of the equation x + y = 6

⇒ y = 6 – x

When x = 2, then y = 6 – 2 = 4 

When x = 3, then y = 6 – 3 = 3

Thus, we have the following table:

x	2	3
y	4	3

Now, plot the points A(2, 4) and B(3, 3) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + y = 6.

Graph of the equation x – y = 2

⇒ y = x – 2

When x = 3, then y = 3 – 2 = 1 

When x = 4, then y = 4 – 2 = 2 

Thus, we have the following table:

x	3	4
y	1	2

Now, plot the points C(3, 1) and D(4, 2) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of x – y = 2.

The two graph lines intersect at point D(4, 2).Question 12

Two students A and B contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation to satisfy the above data and draw its graph.Solution 12

Let the amount contributed by students A and B be Rs. x and Rs. y respectively.

Total contribution = 100

⇒ x + y = 100

⇒ y = 100 – x

When x = 25, then y = 100 – 25 = 75

When x = 50, then y = 100 – 50 = 50

Thus, we have the following table:

x	25	50
y	75	50

Now, plot the points A(25, 75) and B(50, 50) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + y = 100.

Exercise Ex. 4A

Question 1(i)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3x + 5y = 7.5 Solution 1(i)

We have,

3x + 5y = 7.5

⇒ 3x + 5y – 7.5 = 0

⇒ 6x + 10y – 15 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 6, b = 10 and c = -15 Question 1(ii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(ii)

On comparing this equation with ax + by + c = 0, we obtain

a = 10, b = -1 and c = 30 Question 1(iii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3y – 2x = 6Solution 1(iii)

We have,

3y – 2x = 6

⇒ -2x + 3y – 6 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = -2, b = 3 and c = -6 Question 1(iv)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

4x = 5ySolution 1(iv)

We have,

4x = 5y

⇒ 4x – 5y = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 4, b = -5 and c = 0 Question 1(v)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(v)

⇒ 6x – 5y = 30

⇒ 6x – 5y – 30 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 6, b = -5 and c = -30 Question 1(vi)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(vi)

On comparing this equation with ax + by + c = 0, we obtain

a = , b =  and c = -5 Question 2(i)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

x = 6Solution 2(i)

We have,

x = 6

⇒ x – 6 = 0

⇒ 1x + 0y – 6 = 0

⇒ x + 0y – 6 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 1, b = 0 and c = -6 Question 2(ii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3x – y = x – 1Solution 2(ii)

We have,

3x – y = x – 1

⇒ 3x – x – y + 1 = 0

⇒ 2x – y + 1 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 2, b = -1 and c = 1 Question 2(iii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

2x + 9 = 0Solution 2(iii)

We have,

2x + 9 = 0

⇒ 2x + 0y + 9 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 2, b = 0 and c = 9 Question 2(iv)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

4y = 7Solution 2(iv)

We have,

4y = 7

⇒ 0x + 4y – 7 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 0, b = 4 and c = -7 Question 2(v)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

x + y = 4Solution 2(v)

We have,

x + y = 4

⇒ x + y – 4 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 1, b = 1 and c = -4 Question 2(vi)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 2(vi)

We have,

⇒ 3x – 8y – 1 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 3, b = -8 and c = -1 Question 3(i)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(4, 0)Solution 3(i)

Given equation is 5x – 4y = 20

Substituting x = 4 and y = 0 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(4) – 4(0)

= 20 – 0

= 20

= R.H.S.

Hence, (4, 0) is the solution of the given equation.Question 3(ii)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(0, 5)Solution 3(ii)

Given equation is 5x – 4y = 20

Substituting x = 0 and y = 5 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(0) – 4(5)

= 0 – 20

= -20

≠ R.H.S.

Hence, (0, 5) is not the solution of the given equation. Question 3(iii)

Check which of the following are the solutions of the equation 5x – 4y = 20.

Solution 3(iii)

Given equation is 5x – 4y = 20

Substituting x = -2 and y =  in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(-2) – 4

= -10 – 10

= -20

≠ R.H.S.

Hence,  is not the solution of the given equation. Question 3(iv)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(0, -5)Solution 3(iv)

Given equation is 5x – 4y = 20

Substituting x = 0 and y = -5 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(0) – 4(-5)

= 0 + 20

= 20

= R.H.S.

Hence, (0, -5) is the solution of the given equation. Question 3(v)

Check which of the following are the solutions of the equation 5x – 4y = 20.

Solution 3(v)

Given equation is 5x – 4y = 20

Substituting x = 2 and y =  in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(2) – 4

= 10 + 10

= 20

= R.H.S.

Hence,  is the solution of the given equation. Question 4(a)

Find five different solutions of each of the following equations:

2x – 3y = 6Solution 4(a)

Given equation is 2x – 3y = 6

Substituting x = 0 in the given equation, we get

2(0) – 3y = 6

⇒ 0 – 3y = 6

⇒ 3y = -6

⇒ y = -2

So, (0, -2) is the solution of the given equation.

Substituting y = 0 in the given equation, we get

2x – 3(0) = 6

⇒ 2x – 0 = 6

⇒ 2x = 6

⇒ x = 3

So, (3, 0) is the solution of the given equation.

Substituting x = 6 in the given equation, we get

2(6) – 3y = 6

⇒ 12 – 3y = 6

⇒ 3y = 6

⇒ y = 2

So, (6, 2) is the solution of the given equation.

Substituting y = 4 in the given equation, we get

2x – 3(4) = 6

⇒ 2x – 12 = 6

⇒ 2x = 18

⇒ x = 9

So, (9, 4) is the solution of the given equation.

Substituting x = -3 in the given equation, we get

2(-3) – 3y = 6

⇒ -6 – 3y = 6

⇒ 3y = -12

⇒ y = -4

So, (-3, -4) is the solution of the given equation.Question 4(b)

Find five different solutions of each of the following equations:

Solution 4(b)

Given equation is  

Substituting x = 0 in (i), we get

4(0) + 3y = 30

⇒ 3y = 30

⇒ y = 10

So, (0, 10) is the solution of the given equation.

Substituting x = 3 in (i), we get

4(3) + 3y = 30

⇒ 12 + 3y = 30

⇒ 3y = 18

⇒ y = 6

So, (3, 6) is the solution of the given equation.

Substituting x = -3 in (i), we get

4(-3) + 3y = 30

⇒ -12 + 3y = 30

⇒ 3y = 42

⇒ y = 14

So, (-3, 14) is the solution of the given equation.

Substituting y = 2 in (i), we get

4x + 3(2) = 30

⇒ 4x + 6 = 30

⇒ 4x = 24

⇒ x = 6

So, (6, 2) is the solution of the given equation.

Substituting y = -2 in (i), we get

4x + 3(-2) = 30

⇒ 4x – 6 = 30

⇒ 4x = 36

⇒ x = 9

So, (9, -2) is the solution of the given equation.Question 4(c)

Find five different solutions of each of the following equations:

3y = 4xSolution 4(c)

Given equation is 3y = 4x

Substituting x = 3 in the given equation, we get

3y = 4(3)

⇒ 3y = 12

⇒ y = 4

So, (3, 4) is the solution of the given equation.

Substituting x = -3 in the given equation, we get

3y = 4(-3)

⇒ 3y = -12

⇒ y = -4

So, (-3, -4) is the solution of the given equation.

Substituting x = 9 in the given equation, we get

3y = 4(9)

⇒ 3y = 36

⇒ y = 12

So, (9, 12) is the solution of the given equation.

Substituting y = 8 in the given equation, we get

3(8) = 4x

⇒ 4x = 24

⇒ x = 6

So, (6, 8) is the solution of the given equation.

Substituting y = -8 in the given equation, we get

3(-8) = 4x

⇒ 4x = -24

⇒ x = -6

So, (-6, -8) is the solution of the given equation.Question 5

If x = 3 and y = 4 is a solution of the equation 5x – 3y = k, find the value of k.Solution 5

Since x = 3 and y = 4 is a solution of the equation 5x – 3y = k, substituting x = 3 and y = 4 in equation 5x – 3y = k, we get

5(3) – 3(4) = k

⇒ 15 – 12 = k

⇒ k = 3 Question 6

If x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, find the value of k.Solution 6

Since x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, substituting these values in equation, we get

4(3k + 2) – 3(2k – 1) + 1 = 0

⇒ 12k + 8 – 6k + 3 + 1 = 0

⇒ 6k + 12 = 0

⇒ 6k = -12

⇒ k = -2 Question 7

The cost of 5 pencils is equal to the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y).Solution 7

Let the cost of one pencil be Rs. x and that of one ballpoint be Rs. y.

Then,

Cost of 5 pencils = Rs. 5x

Cost of 2 ballpoints = Rs. 2y

According to given statement, we have

5x = 2y

⇒ 5x – 2y = 0